Carleman's inequality

Remarks

Proof idea: let $c_k = (k + 1)^k / k^{k-1}$, so that $c_1 c_2 \cdots c_n = (n + 1)^n$. By the AM–GM inequality, $$(a_1 \cdots a_n)^{1/n} = \frac{(c_1 a_1 \cdots c_n a_n)^{1/n}}{n + 1} \le \frac{1}{n(n+1)} \sum_{k=1}^{n} c_k a_k.$$ Summing over $n$ and using $\sum_{n \ge k} \frac{1}{n(n+1)} = \frac{1}{k}$ gives the bound $\sum_k (1 + 1/k)^k a_k$, and $(1 + 1/k)^k < e$.

The constant $e$ cannot be replaced by a smaller number. If $0 < \sum a_n < \infty$, the inequality is strict.

Going deeper

For a student

Carleman's inequality is a limiting case of the discrete Hardy's inequality. Apply Hardy's inequality to $a_n^{1/p}$ and let $p \to \infty$: the power means tend to geometric means, and $\big(\frac{p}{p-1}\big)^p \to e$. The direct proof above shows how a well chosen weighting makes the AM–GM inequality sharp enough.